Question: What is the value of $\dfrac{d}{dx}\cot(x)$ at $x=\dfrac{3\pi}{2}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $-1$ (Choice C) C $\dfrac12$ (Choice D) D $0$
Answer: Let's first find $\dfrac{d}{dx}\cot(x)$. Then, we can evaluate it at $x=\dfrac{3\pi}{2}$. Recall that the derivative of $\cot(x)$ is $-\dfrac{1}{\sin^2(x)}$, or $-\csc^2(x)$. Put another way, $\dfrac{d}{dx}[\cot(x)]=-\dfrac{1}{\sin^2(x)}=-\csc^2(x)$. [Is there a way to know this without memorizing?] Now let's plug in $x={\dfrac{3\pi}{2}}$ : $\begin{aligned} &\phantom{=}-\dfrac{1}{\sin^2\left({\dfrac{3\pi}{2}}\right)} \\\\ &=-\dfrac{1}{(-1)^2} \\\\ &=-1 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\cot(x)$ at $x=\dfrac{3\pi}{2}$ is $-1$.